Question

The sum of all possible real values of $x$ for which \[ \log_{x-3}(x^2 - 9) = \log_{x-3}(x + 1) + 2, \] is

• A. \(-3\)
• B. \(\sqrt{33}\)
• C. \(\dfrac{3 + \sqrt{33}}{2}\)
• D. \(3\)

Hint:

In logarithmic equations:
Always check the domain first: base $> 0$, base $\neq 1$, and argument $> 0$.
When the bases are the same, combine logs using properties like $\log_b A - \log_b B = \log_b \left(\dfrac{A}{B}\right)$, then convert to exponential form.
Don’t forget to discard any solutions that fall outside the domain constraints.

Answer 1

The Correct Option is C

Approach: Convert to one log on each side and eliminate the matching base, but this time keep the equation as a cubic and read off which roots the domain throws away $-$ that's where the wording "sum of all possible values" gets tested.

Step 1: Using $a=\log_{x-3}(x-3)$, the $+2$ on the right is $\log_{x-3}(x-3)^2$. Equate arguments: $x^2-9=(x+1)(x-3)^2.$

Step 2: Expand fully. $(x+1)(x-3)^2=(x+1)(x^2-6x+9)=x^3-5x^2+3x+9.$ So $x^2-9=x^3-5x^2+3x+9$, giving $x^3-6x^2+3x+18=0.$

Step 3: Factor by grouping: $x^2(x-6)+3(x-6)=(x-6)(x^2+3)$... that's not matching, so test $x=3$: $27-54+9+18=0.$ Yes, so $(x-3)$ is a factor. Dividing: $x^3-6x^2+3x+18=(x-3)(x^2-3x-6).$

Step 4: Roots are $x=3$ and $x=\dfrac{3\pm\sqrt{33}}{2}.$ Domain needs $x>3$ (base positive, $\neq1$). Reject $x=3$ (base $0$) and $x=\dfrac{3-\sqrt{33}}{2}\approx-1.37$. Only $\dfrac{3+\sqrt{33}}{2}$ qualifies.

Final answer: Sum of valid values $=\dfrac{3+\sqrt{33}}{2}.$